
\prob{006D}{指数相等}

若自然数$a, b$满足$a^x = b^y = 1994^z$，同时有$\sfrac1x + \sfrac1y = \sfrac1z$，求$a, b$的所有可能值。
\problabels{yellow/数论, green/代数求值问题}

\ans{$a = 1, b = 1994$；$a = 2, b = 997$；$a = 997, b = 2$；$a = 1994, b = 1$。}

\subsection{对数}

基本思路：等式求对数，然后通过代数变换化简得$\ln a + \ln b = \ln 1994$，即$ab = 1994$。

等式求自然对数得

\[ x\ln a = y\ln b = z\ln1994 \]

由$\sfrac1x + \sfrac1y = \sfrac1z$知

\[ z = \frac{xy}{x + y} \]

故

\begin{align*}
  x\ln a = y\ln b &= \frac{xy}{x + y}\ln1994 \\
  x(x + y)\ln a = y(x + y)\ln b &= xy\ln1994 \\
\end{align*}

于是有

\begin{align*}
  (x + y)\ln a &= y\ln1994 \\
  (x + y)\ln b &= x\ln1994 \\
  (x + y)(\ln a + \ln b) &= (x + y)\ln1994 \\
  \ln ab &= \ln1994 \\
  ab &= 1994 \\
\end{align*}

而$a, b$是自然数，故有$4$种情况：$a = 1, b = 1994$；$a = 2, b = 997$；$a = 997, b = 2$；$a = 1994, b = 1$。
